Equating the coefficients of the matrices in the last two lines gives\[ \left\{\begin{array}{rl} a=&a+dp, \\ b=& b\sqrt{1-p}, \\ c=& c\sqrt{1-p}, \\ d=&d(1-p) \\ \end{array} \right.\] Since $0<p<1$, $1-p\neq 0,1$, and thus $b=c=d=0$. Furthermore, since $\rho_0$ is a density matrix, $Tr(\rho_0)=a+d=a=1$. Thus,\[ \rho_0=\begin{pmatrix} 1&0\\0&0\end{pmatrix}=\ket{0}\bra{0}\] is the fixed point satisfying $D_p(\rho_0)=\rho_0$.

Now, lets show that the operation $D^{(2)}_p$, corresponding to applying $D_p$ twice in succession, that is $D^{(2)}_p(\rho)=D_p(D_p(\rho))$, is equivalent to applying $D_q$ once for some suitably chosen value $q$.

Generalizing the case just analyzed, we can also define the operation $D^{(k)}_p$, corresponding to applying $D_p$ $k$ times in succession. What are the Krauss operators of $D^{(k)}_p$ with their matrix entries written as closed-form expressions in terms of $p$ and $k$?

Calculating the effect of applying $D_p(\rho)$ $k-$times in succession results in the expression\[ D_p^{(k)}(\rho)=A_0^kA_0^{k\dagger}+\SUM{n=1}{k}(A_1A_0^{n-1})\rho\SUM{n=1}{k}(A_1A_0^{n-1})^\dagger.\]

Let \[ B_0^{(k)}:=A_0^kA_0^{k\dagger}=\begin{pmatrix}1&0\\0&(1-p)^{k/2} \end{pmatrix} \ \ \text{and} \ \ B_1^{(k)}=\SUM{n=1}{k}(A_1A_0^{n-1})=\begin{pmatrix}a&b\\c&d\end{pmatrix}\] be the two Krauss operators of $D_p^{(k)}$ (here, $B_1^{(k)}$ has been expressed as some matrix with unknown coefficients which are to be determined) so that\[ D_p^{(k)}(\rho)= B_0^{(k)}\rho B_0^{(k)\dagger}+ B_1^{(k)} \rho B_1^{(k)\dagger}.\] Since $D_p$ itself is a valid quantum operation, compositions of $D_{p}$ will also always be another valid quantum operation. Thus, the two Krauss operators for $D_p^{(k)}$ must satisfy the relation $I=B_0^{(k)\dagger} B_0^{(k)}+ B_1^{(k)^\dagger} B_1^{(k)}$. Thus. $B_1^{(k)}$ can be determined more precisely by the identity $ B_1^{(k)\dagger}B_1^{(k)}=I-B_0^{(k)^\dagger}B_0^{(k)}$. Since\[ B_0^{(k)^\dagger}B_0^{(k)}=\begin{pmatrix}1&0\\0&(1-p)^k\end{pmatrix} \ \ \text{and} \ \ B_1^{(k)\dagger}B_1^{(k)}= \begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}\]

then using the previous relation implies\[\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}=\begin{pmatrix}0&0\\0&1-(1-p)^{k} \end{pmatrix}.\]

From this, it is seen that\[ \begin{align*} a^2+c^2=0 &\implies a=c=0 \\ b^2+d^2&=1-(1-p)^{k}, \end{align*}\] but in the case when $k=1$, $B_1^{(1)}=A_1$ so that $b=\sqrt{p}$ and $b^2+d^2=1-(1-p)=p$ implying that $d=0$. Hence, in general\[ B_1^{(k)}=\begin{pmatrix}0&\sqrt{1-(1-p)^k}\\0&0\end{pmatrix} \ \ \text{and} \ \ B_0^{(k)}=\begin{pmatrix}1&0\\0&(1-p)^{k/2} \end{pmatrix}\] are the Krauss operators of $D_p^{(k)}$.

Now, lets answer the following question: Is $lim_{k\to \infty}D^(k)_p(\rho)=\rho_0$ for any initial state $\rho$, where $\rho_0$ is the fixed point of $D_p$ that was calculated in part above?

For $0<p<1$, it is always the case that $lim_{k\to\infty}(1-p)^k=0$. Therefore, in the limit,\[ lim_{k\to\infty}B_0^{(k)}=\begin{pmatrix}1&0\\0&0 \end{pmatrix} \ \ \text{and} \ \ lim_{k\to\infty}B_1^{(k)}=\begin{pmatrix}0&1\\0&0 \end{pmatrix}.\] Let $\rho$ be an arbitrary density matrix expressed in the general form\[\rho=\begin{pmatrix}a&b\\c&d \end{pmatrix},\]where the constraint $a+d=1$ is imposed so that $Tr(\rho)=1$ as required for density operators by definition. Then it follows that\[\begin{align*}lim_{k\to \infty}D^(k)_p(\rho)&=\begin{pmatrix}1&0\\0&0 \end{pmatrix}\rho\begin{pmatrix}1&0\\0&0 \end{pmatrix}+\begin{pmatrix}0&1\\0&0 \end{pmatrix}\rho\begin{pmatrix}0&0\\1&0 \end{pmatrix} =\begin{pmatrix}a+d&0\\0&0 \end{pmatrix}=\begin{pmatrix}1&0\\0&0 \end{pmatrix}=\rho_0,\end{align*}\]which is precisely equal to the fixed point calculated in above.